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Determine the boundary work done to compress 3 kg of nitrogen gas in a piston-cylinder device at 27°C, 0.15 MPa isothermally to 0.3 MPa. Express you answer in kJ to 5 significant figures.

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Step-by-step explanation:

The boundary work done when compressing a gas is given by the equation:

Work = P₁ * V₁ * ln(V₂/V₁)

Where:

P₁ = Initial pressure

V₁ = Initial volume

V₂ = Final volume

In this case, you have:

Initial pressure (P₁) = 0.15 MPa = 0.15 * 10^6 Pa

Final pressure (P₂) = 0.3 MPa = 0.3 * 10^6 Pa

Initial volume (V₁) is not given, but it's not needed for this calculation.

Final volume (V₂) is also not given, but we can find it using the ideal gas law.

You can find the final volume using the ideal gas law:

PV = nRT

Where:

P = pressure

V = volume

n = number of moles

R = ideal gas constant

T = temperature (in Kelvin)

First, we need to find the number of moles (n). Since you have 3 kg of nitrogen gas:

n = (mass) / (molar mass)

The molar mass of nitrogen (N2) is approximately 28.02 g/mol, so:

n = (3 kg) / (0.02802 kg/mol) = 107.11 moles

Now, you can find the final volume (V₂) at the new pressure (P₂) and the same temperature (T):

V₂ = (nRT) / P₂

T = 27°C = 273 K (convert to Kelvin)

V₂ = (107.11 moles * 8.314 J/(mol·K) * 273 K) / (0.3 * 10^6 Pa) = 789.87 liters = 0.78987 m³

Now that you have V₂, you can calculate the boundary work:

Work = P₁ * V₁ * ln(V₂/V₁)

Work = (0.15 * 10^6 Pa) * V₁ * ln(0.78987 m³ / V₁)

To isolate V₁, you can solve for it:

ln(0.78987 / V₁) = Work / ((0.15 * 10^6 Pa) * V₁)

Now, take the natural logarithm of both sides:

0.78987 / V₁ = e^(Work / ((0.15 * 10^6 Pa) * V₁))

To solve for V₁, you can rearrange the equation:

V₁ = 0.78987 / e^(Work / ((0.15 * 10^6 Pa) * V₁))

At this point, you'll need to use a numerical method or software to solve for V₁ because it's an implicit equation. Once you find V₁, you can calculate the work done.

The work done in this isothermal compression process is equal to the negative of the area under the pressure-volume curve, but finding the exact value would require numerical calculations or a software tool.

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